Falling elevator onto a spring.

• J-dizzal
In summary: J is the work done on the elevator before the spring is compressed... But as the spring compresses, gravity and friction still act on the elevator (doing additional work).Your mistake is basically the same as your mistake in post #4 of your previous... you are forgetting to account for the work done by gravity and friction as the spring compresses.In summary, the elevator cab with a weight of 2000 kg and a constant frictional force of 2.4 kN experiences a snapped cable while at rest on the first floor, where the cab bottom is 2.8 m above a spring with a spring constant of
J-dizzal

Homework Statement

The cable of the 2000 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 2.8 m above a spring of spring constant k = 0.30 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 2.4 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_59.jpg

The Attempt at a Solution

I got part a, but part b is where I am stuck. i got an answer of .0573m I am not sure how to find distance without using the spring equation.
[/B]

x=(mg-Ff)/k is when the acceleration is zero, but you want the velocity to be zero not acceleration.

If the spring is compressed by an amount x, then how much work from gravity and friction is done? And if the velocity is zero, then what is the spring-potential-energy?

Nathanael said:
x=(mg-Ff)/k is when the acceleration is zero, but you want the velocity to be zero not acceleration.

If the spring is compressed by an amount x, then how much work from gravity and friction is done? And if the velocity is zero, then what is the spring-potential-energy?
x amount would be w=1/2 kx2 + 48160J.
If velocity is zero then spring potential energy is at the maximum.
Edit
work done = 17200d
max being = 48160J

Last edited:
J-dizzal said:
x amount would be w=1/2 kx2 + 48160J.
If velocity is zero then spring potential energy is at the maximum.
Edit
work done = 17200d
max being = 48160J
Gravity and friction still do work while the spring is being compressed, so it should depend on x.

Nathanael said:
Gravity and friction still do work while the spring is being compressed, so it should depend on x.
would work be (19600+2400)x? because the spring is working against friction? But so is the gravity force.

J-dizzal said:
would work be (19600+2400)x? because the spring is working against friction? But so is the gravity force.
I'm not sure what you mean by 'the spring works against friction.' Gravity pulls it down (positive work since it's falling) and the spring and friction push it up (negative work).

You had the work done by gravity and friction almost correct before when you put (mg-Ff)d but "d" is not just the distance above the spring, it also depends on how much the spring is compressed.

Nathanael said:
I'm not sure what you mean by 'the spring works against friction.' Gravity pulls it down (positive work since it's falling) and the spring and friction push it up (negative work).

You had the work done by gravity and friction almost correct before when you put (mg-Ff)d but "d" is not just the distance above the spring, it also depends on how much the spring is compressed.
work=Fd=17200(2.8)=48160J.
48160=1/2 kx2. but when I solve for x i get .566 its wrong

J-dizzal said:
work=Fd=17200(2.8)=48160J.
48160=1/2 kx2. but when I solve for x i get .566 its wrong
d is not 2.8, d depends on x

Nathanael said:
d is not 2.8, d depends on x
d=2.8m was given. I am using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.

J-dizzal said:
d=2.8m was given. I am using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.
2.8m is given as the distance between the spring and the bottom of the elevator... But the elevator falls farther than that doesn't it? If the elevator only fell 2.8m then the spring wouldn't be compressed.

Nathanael said:
2.8m is given as the distance between the spring and the bottom of the elevator... But the elevator falls farther than that doesn't it? If the elevator only fell 2.8m then the spring wouldn't be compressed.
That is the distance I am trying to find. I was thinking if the energy of the elevator hits the spring with 48160J the spring must compress and equal that energy when it comes to a stop at distance x in the formula W= 1/2 kx2

J-dizzal said:
d=2.8m was given. I'm using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.
Adding to what Nathanael posted ...

You have found the work done by gravity and friction in falling prior to contact with the spring. How much work do gravity and friction do over the distance, x, when the spring is being compressed?

(I would have thought that you would have these spring problems licked by now.)

SammyS said:
Adding to what Nathanael posted ...

You have found the work done by gravity and friction in falling prior to contact with the spring. How much work do gravity and friction do over the distance, x, when the spring is being compressed?

(I would have thought that you would have these spring problems licked by now.)
these problems are licking me.
when the spring is being compressed it does a total of -48160J? because its oppose of the KE

edit; or would it be w=17200x where x is the distance the spring is compressed.

J-dizzal said:
when the spring is being compressed it does a total of -48160J? because its oppose of the KE
48160J is the work done on the elevator before the spring is compressed... But as the spring compresses, gravity and friction still act on the elevator (doing additional work).

J-dizzal said:
these problems are licking me.
when the spring is being compressed it does a total of -48160J? because its oppose of the KE

edit; or would it be w=17200x where x is the distance the spring is compressed.
BINGO!

It's the 17200x which you have been neglecting and Nathanael has been hinting at strongly.

Nathanael said:
48160J is the work done on the elevator before the spring is compressed... But as the spring compresses, gravity and friction still act on the elevator (doing additional work).
.30M N/M = 300000NM?

J-dizzal said:
.30M N/M = 300000NM?
Yes, 0.3 Mega N/m is 300,000 N/m

SammyS
SammyS said:
BINGO!

It's the 17200x which you have been neglecting and Nathanael has been hinting at strongly.
but then i get 2.8m again
48160 +17200x = W, solve for x its 2.8m

J-dizzal said:
but then i get 2.8m again
48160 +17200x = W, solve for x its 2.8m
By "W" you mean the work done by the spring, right? You should get a quadratic equation.

Nathanael said:
By "W" you mean the work done by the spring, right? You should get a quadratic equation.
W=1/2 k (48160+17200x)x I am really confused.

J-dizzal said:
W=1/2 k (48160+17200x)x I am really confused.
I agree. You're really confused.

Read the last line of Post #11 .

SammyS said:
I agree. You're really confused.

Read the last line of Post #11 .
was that the correct equation to you?

J-dizzal said:
was that the correct equation to you?
so W=1/2 k x2 I am not sure how 17200x fits into that

The spring's potential energy is 0.5kx2 and (since the final speed is zero) this is equal to the work done by gravity&friction, which you've found is equal to 17200N times the distance the elevator falls.

Everything in the above sentence makes complete sense right?

So the million dollar question is, how far does the elevator fall? Hint: the answer is not 2.8m

Nathanael said:
The spring's potential energy is 0.5kx2 and (since the final speed is zero) this is equal to the work done by gravity&friction, which you've found is equal to 17200N times the distance the elevator falls.

Everything in the above sentence makes complete sense right?

So the million dollar question is, how far does the elevator fall? Hint: the answer is not 2.8m
why 17200N what is the N?

J-dizzal said:
why 17200N what is the N?
17200N=1/2 kx2 Sorry I am not getting it.

J-dizzal said:
why 17200N what is the N?
The N just means Newtons. I included it because the equations don't actually have to use SI units to get the correct answer, and the number 17200 is meaningless without a unit. You should really get in the habit of not plugging in numbers until you're finished with the problem (or at least almost finished, because sometimes it gets messy). So instead of writing 17200, I would write (mg-Ff).

Nathanael said:
The N just means Newtons. I included it because the equations don't actually have to use SI units to get the correct answer, and the number 17200 is meaningless without a unit. You should really get in the habit of not plugging in numbers until you're finished with the problem (or at least almost finished, because sometimes it gets messy). So instead of writing 17200, I would write (mg-Ff).
mg-Ff = 1/2 kx2

J-dizzal said:
mg-Ff = 1/2 kx2
Does this equation mean something to you?

The units are not even the same on both sides. Force is never equal to energy.

Look at the bigger picture and take it slow.

Nathanael said:
Does this equation mean something to you?

The units are not even the same on both sides. Force is never equal to energy.

Look at the bigger picture and take it slow.
(17200N)x=1/2 300000N/m
x=8.721m not sure what I am doing wrong.

edit. i can't put an x2 on the right side because then i cannot solve for x

J-dizzal said:
(17200N)x=1/2 300000N/m
x=8.721m not sure what I am doing wrong.
The units are not even right again.

J-dizzal said:
edit. i can't put an x2 on the right side because then i cannot solve for x
First, yes you could still solve for x
Second, you can't just leave out an important part of an equation because it makes it harder to solve!
We are making no progress. Let us start from square one...
Please explain to me your approach to this problem. How are you going to find where the velocity is zero?

if the energy of the elevator hits the spring with 48160J how far does the spring compress? 17200x =48160?

Nathanael said:
The units are not even right again.First, yes you could still solve for x
Second, you can't just leave out an important part of an equation because it makes it harder to solve!
We are making no progress. Let us start from square one...
Please explain to me your approach to this problem. How are you going to find where the velocity is zero?
velocity =0 when the spring distance is maximum, or when the work done by the spring = 48160 +17200x where x is the distance of spring compression.

J-dizzal said:
velocity =0 ... when the work done by the spring = 48160 +17200x where x is the distance of spring compression.
Yes, exactly. Now what is the work done by the spring?

Nathanael said:
Yes, exactly. Now what is the work done by the spring?
48160J +17200x = Wspring

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